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Quantitative Analysis for Management

Thirteenth Edition

Lesson 4

Decision Analysis

(Based on Chapter 3)

Learning Objectives

After completing this lesson, students will be able to:

3.1 List the steps of the decision-making process.

3.2 Describe the types of decision-making environments.

3.3 Make decisions under uncertainty.

3.4 Use probability values to make decisions under risk.

3.5 Use computers to solve basic decision-making problems.

3.6 Develop accurate and useful decision trees.

Lesson Outline

3.1 The Six Steps in Decision Making

3.2 Types of Decision-Making Environments

3.3 Decision Making Under Uncertainty

3.4 Decision Making Under Risk

3.5 Using Software for Payoff Table Problems

3.6 Decision Trees

Introduction

What is involved in making a good decision?

Decision theory is an analytic and systematic approach to the study of decision making

A good decision is one that is based on logic, considers all available data and possible alternatives, and applies a quantitative approach

The Six Steps in Decision Making

Clearly define the problem at hand

List the possible alternatives

Identify the possible outcomes or states of nature

List the payoff (typically profit) of each combination of alternatives and outcomes

Select one of the mathematical decision theory models

Apply the model and make your decision

Thompson Lumber Company (1 of 3)

Step 1 – Define the problem

Consider expanding by manufacturing and marketing a new product – backyard storage sheds

Step 2 – List alternatives

Construct a large new plant

Construct a small new plant

Do not develop the new product line

Step 3 – Identify possible outcomes, states of nature

The market could be favorable or unfavorable

Thompson Lumber Company (2 of 3)

Step 4 – List the payoffs

Identify conditional values for the profits for large plant, small plant, and no development for the two possible market conditions

Step 5 – Select the decision model

Depends on the environment and amount of risk and uncertainty

Step 6 – Apply the model to the data

Thompson Lumber Company (3 of 3)

TABLE 3.1 Decision Table with Conditional Values for Thompson Lumber

STATE OF NATURE

Blank	FAVORABLE MARKET	UNFAVORABLE MARKET
ALTERNATIVE	($)	($)
Construct a large plant	200,000	−180,000
Construct a small plant	100,000	−20,000
Do nothing	0	0

Note: It is important to include all alternatives, including “do nothing.”

Types of Decision-Making Environments

Decision making under certainty

The decision maker knows with certainty the consequences of every alternative or decision choice

Decision making under uncertainty

The decision maker does not know the probabilities of the various outcomes

Decision making under risk

The decision maker knows the probabilities of the various outcomes

Decision Making Under Uncertainty

Criteria for making decisions under uncertainty

Maximax (optimistic)

Maximin (pessimistic)

Criterion of realism (Hurwicz)

Equally likely (Laplace)

Minimax regret

Optimistic

Used to find the alternative that maximizes the maximum payoff – maximax criterion

Locate the maximum payoff for each alternative

Select the alternative with the maximum number

TABLE 3.2 Thompson’s Maximax Decision

STATE OF NATURE

Blank	FAVORABLE	UNFAVORABLE	MAXIMUM IN
Blank	MARKET	MARKET	A ROW
ALTERNATIVE	($)	($)	($)
Construct a large plant	200,000	−180,000	200,000
Blank	Blank	Blank	Maximax
Construct a small plant	100,000	−20,000	100,000
Do nothing	0	0	0

Pessimistic

Used to find the alternative that maximizes the minimum payoff – maximin criterion

Locate the minimum payoff for each alternative

Select the alternative with the maximum number

TABLE 3.3 Thompson’s Maximin Decision

STATE OF NATURE

Blank	FAVORABLE	UNFAVORABLE	MAXIMUM IN
Blank	MARKET	MARKET	A ROW
ALTERNATIVE	($)	($)	($)
Construct a large plant	200,000	−180,000	−180,000
Construct a small plant	100,000	−20,000	−20,000
Do nothing	0	0	0
Blank	Blank	Blank	Maximin

Criterion of Realism (Hurwicz) (1 of 2)

Often called weighted average

Compromise between optimism and pessimism

Select a coefficient of realism α, with 0 ≤ α ≤ 1

α = 1 is perfectly optimistic

α = 0 is perfectly pessimistic

Compute the weighted averages for each alternative

Select the alternative with the highest value

Weighted average = α(best in row)

+ (1−α)(worst in row)

Criterion of Realism (Hurwicz) (2 of 2)

For the large plant alternative using α = 0.8

(0.8)(200,000) + (1−0.8)(−180,000) = 124,000

For the small plant alternative using α = 0.8

(0.8)(100,000) + (1−0.8)(−20,000) = 76,000

TABLE 3.4 Thompson’s Criterion of Realism Decision

STATE OF NATURE

Blank	FAVORABLE	UNFAVORABLE	CRITERION OF REALISM
Blank	MARKET	MARKET	OR WEIGHTED AVERAGE
ALTERNATIVE	($)	($)	(α = 0.8) ($)
Construct a large plant	200,000	−180,000	124,000
Blank	Blank	Blank	Realism
Construct a small plant	100,000	−20,000	76,000
Do nothing	0	0	0

Equally Likely (Laplace)

Considers all the payoffs for each alternative

Find the average payoff for each alternative

Select the alternative with the highest average

TABLE 3.5 Thompson’s Equally Likely Decision

STATE OF NATURE

Blank	FAVORABLE	UNFAVORABLE	Blank
Blank	MARKET	MARKET	ROW AVERAGE
ALTERNATIVE	($)	($)	($)
Construct a large plant	200,000	−180,000	10,000
Construct a small plant	100,000	−20,000	40,000
Blank	Blank	Blank	Equally likely
Do nothing	0	0	0

Minimax Regret (1 of 4)

Based on opportunity loss or regret

The difference between the optimal profit and actual payoff for a decision

Create an opportunity loss table by determining the opportunity loss from not choosing the best alternative

Calculate opportunity loss by subtracting each payoff in the column from the best payoff in the column

Find the maximum opportunity loss for each alternative and pick the alternative with the minimum number

Minimax Regret (2 of 4)

TABLE 3.6 Determining Opportunity Losses for Thompson Lumber

STATE OF NATURE

FAVORABLE	UNFAVORABLE
MARKET	MARKET
($)	($)
200,000 − 200,000	0 − (−180,000)
200,000 − 100,000	0 − (−20,000)
200,000 − 0	0 − 0

Minimax Regret (3 of 4)

TABLE 3.7 Opportunity Loss Table for Thompson Lumber

STATE OF NATURE

Blank	FAVORABLE	UNFAVORABLE
Blank	MARKET	MARKET
ALTERNATIVE	($)	($)
Construct a large plant	0	180,000
Construct a small plant	100,000	20,000
Do nothing	200,000	0

Minimax Regret (4 of 4)

TABLE 3.8 Thompson’s Minimax Decision Using Opportunity Loss

STATE OF NATURE

Blank	FAVORABLE	UNFAVORABLE	MAXIMUM IN
Blank	MARKET	MARKET	A ROW
ALTERNATIVE	($)	($)	($)
Construct a large plant	0	180,000	180,000
Construct a small plant	100,000	20,000	100,000
Blank	Blank	Blank	Minimax
Do nothing	200,000	0	200,000

Decision Making Under Risk (1 of 2)

When there are several possible states of nature and the probabilities associated with each possible state are known

Most popular method – choose the alternative with the highest expected monetary value (EMV)

where

Xi = payoff for the alternative in state of nature i

P(Xi) = probability of achieving payoff Xi (i.e., probability of state of nature i)

∑ = summation symbol

Decision Making Under Risk (2 of 2)

Expanding the equation

EMV (alternative i) = (payoff of first state of nature)

×(probability of first state of nature)

+ (payoff of second state of nature)

×(probability of second state of nature)

+ … + (payoff of last state of nature)

×(probability of last state of nature)

EMV for Thompson Lumber (1 of 2)

Each market outcome has a probability of occurrence of 0.50

Which alternative would give the highest EMV?

EMV (large plant) = ($200,000)(0.5) + (−$180,000)(0.5)

= $10,000

EMV (small plant) = ($100,000)(0.5) + (−$20,000)(0.5)

= $40,000

EMV (do nothing) = ($0)(0.5) + ($0)(0.5)

= $0

EMV for Thompson Lumber (2 of 2)

TABLE 3.9 Decision Table with Probabilities and EMVs for Thompson Lumber

STATE OF NATURE

Blank	FAVORABLE	UNFAVORABLE	Blank
Blank	MARKET	MARKET	Blank
ALTERNATIVE	($)	($)	EMV ($)
Construct a large plant	200,000	−180,000	10,000
Construct a small plant	100,000	−20,000	40,000
Blank	Blank	Blank	Best EMV
Do nothing	0	0	0
Probabilities	0.50	0.50	Blank

Expected Value of Perfect Information (EVPI) (1 of 6)

EVPI places an upper bound on what you should pay for additional information

EVwPI is the long run average return if we have perfect information before a decision is made

EVwPI = ∑(best payoff in state of nature i)

(probability of state of nature i)

Expected Value of Perfect Information (EVPI) (2 of 6)

Expanded EVwPI becomes

EVwPI = (best payoff for first state of nature)

× (probability of first state of nature)

+ (best payoff for second state of nature)

× (probability of second state of nature)

+ … + (best payoff for last state of nature)

× (probability of last state of nature)

And

EVPI = EVwPI − Best EMV

Expected Value of Perfect Information (EVPI) (3 of 6)

Scientific Marketing, Inc. offers analysis that will provide certainty about market conditions (favorable)

Additional information will cost $65,000

Should Thompson Lumber purchase the information?

Expected Value of Perfect Information (EVPI) (4 of 6)

TABLE 3.10 Decision Table with Perfect Information

STATE OF NATURE

Blank	FAVORABLE	UNFAVORABLE	Blank
Blank	MARKET	MARKET	Blank
ALTERNATIVE	($)	($)	EMV ($)
Construct a large plant	200,000	−180,000	10,000
Construct a small plant	100,000	−20,000	40,000
Do nothing	0	0	0
With perfect information	200,000	0	100,000
Blank	Blank	Blank	EVwPI
Probabilities	0.50	0.50	Blank

Expected Value of Perfect Information (EVPI) (5 of 6)

The maximum EMV without additional information is $40,000

Therefore

EVPI = EVwPI − Maximum EMV

= $100,000 − $40,000

= $60,000

So the maximum Thompson should pay for the additional information is $60,000

Expected Value of Perfect Information (EVPI) (6 of 6)

The maximum EMV without additional information is $40,000

Therefore

EVPI = EVwPI − Maximum EMV

= $100,000 − $40,000

= $60,000

Thompson should not pay $65,000 for this information

So the maximum Thompson should pay for the additional information is $60,000

Expected Opportunity Loss (1 of 2)

Expected opportunity loss (EOL) is the cost of not picking the best solution

Construct an opportunity loss table

For each alternative, multiply the opportunity loss by the probability of that loss for each possible outcome and add these together

Minimum EOL will always result in the same decision as maximum EMV

Minimum EOL will always equal EVPI

Expected Opportunity Loss (2 of 2)

EOL (large plant) = (0.50)($0) + (0.50)($180,000) = $90,000

EOL (small plant) = (0.50)($100,000) + (0.50)($20,000) = $60,000

EOL (do nothing) = (0.50)($200,000) + (0.50)($0) = $100,000

TABLE 3.11 EOL Table for Thompson Lumber

STATE OF NATURE

Blank	FAVORABLE	UNFAVORABLE	Blank
Blank	MARKET	MARKET	Blank
ALTERNATIVE	($)	($)	EOL ($)
Construct a large plant	0	180,000	90,000
Construct a small plant	100,000	20,000	60,000
Blank	Blank	Blank	Best EOL
Do nothing	200,000	0	100,000
Probabilities	0.50	0.50	Blank

Sensitivity Analysis (1 of 4)

Define P = probability of a favorable market

EMV(large plant) = $200,000P − $180,000)(1 − P)

= $200,000P − $180,000 + $180,000P

= $380,000P − $180,000

EMV(small plant) = $100,000P − $20,000)(1 − P)

= $100,000P − $20,000 + $20,000P

= $120,000P − $20,000

EMV(do nothing) = $0P + 0(1 − P)

= $0

Sensitivity Analysis (2 of 4)

FIGURE 3.1 Sensitivity Analysis

Sensitivity Analysis (3 of 4)

Point 1: EMV(do nothing) = EMV(small plant)

Point 2: EMV(small plant) = EMV(large plant)

Sensitivity Analysis (4 of 4)

FIGURE 3.1 Sensitivity Analysis

BEST ALTERNATIVE	RANGE OF P VALUES
Do nothing	Less than 0.167
Construct a small plant	0.167 − 0.615
Construct a large plant	Greater than 0.615

A Minimization Example (1 of 8)

Three year lease for a copy machine

Which machine should be selected?

TABLE 3.12 Payoff Table with Monthly Copy Costs for Business Analytics Department

Blank	10,000 COPIES PER MONTH	20,000 COPIES PER MONTH	30,000 COPIES PER MONTH
Machine A	950	1,050	1,150
Machine B	850	1,100	1,350
Machine C	700	1,000	1,300

A Minimization Example (2 of 8)

Three year lease for a copy machine

Which machine should be selected?

TABLE 3.13 Best and Worst Payoffs (Costs) for Business Analytics Department

Blank	10,000 COPIES PER MONTH	20,000 COPIES PER MONTH	30,000 COPIES PER MONTH	BEST PAYOFF (MINIMUM)	WORST PAYOFF (MAXIMUM)
Machine A	950	1,050	1,150	950	1,150
Machine B	850	1,100	1,350	850	1,350
Machine C	700	1,000	1,300	700	1,300

A Minimization Example (3 of 8)

Using Hurwicz criteria with 70% coefficient

Weighted average = 0.7(best payoff)

+ (1 − 0.7)(worst payoff)

For each machine

Machine A: 0.7(950) + 0.3(1,150) = 1,010

Machine B: 0.7(850) + 0.3(1,350) = 1,000

Machine C: 0.7(700) + 0.3(1,300) = 880

A Minimization Example (4 of 8)

For equally likely criteria

For each machine

Machine A: (950 + 1,050 + 1,150)÷3 = 1,050

Machine B: (850 + 1,100 + 1,350)÷3 = 1,100

Machine C: (700 + 1,000 + 1,300)÷3 = 1,000

A Minimization Example (5 of 8)

For EMV criteria

USAGE	PROBABILITY
10,000	0.40
20,000	0.30
30,000	0.30

A Minimization Example (6 of 8)

For EMV criteria

TABLE 3.14 Expected Monetary Values and Expected Values with Perfect Information for Business Analytics Department

Blank	10,000 COPIES PER MONTH	20,000 COPIES PER MONTH	30,000 COPIES PER MONTH	EMV
Machine A	950	1,050	1,150	1,040
Machine B	850	1,100	1,350	1,075
Machine C	700	1,000	1,300	970
With perfect information	700	1,000	1,150	925
Probability	0.4	0.3	0.3	Blank

A Minimization Example (7 of 8)

For EVPI

TABLE 3.14 Expected Monetary Values and Expected Values with Perfect Information for Business Analytics Department

Blank	10,000 COPIES PER MONTH	20,000 COPIES PER MONTH	30,000 COPIES PER MONTH	EMV
Machine A	950	1,050	1,150	1,040
Machine B	850	1,100	1,350	1,075
Machine C	700	1,000	1,300	970
With perfect information	700	1,000	1,150	925
Probability	0.4	0.3	0.3	Blank

EVwPI = $925

Best EMV without perfect information = $970

EVPI = 970 − 925 = $45

A Minimization Example (8 of 8)

Opportunity loss criteria

TABLE 3.15 Opportunity Loss Table for Business Analytics Department

Blank	10,000 COPIES PER MONTH	20,000 COPIES PER MONTH	30,000 COPIES PER MONTH	MAXIMUM	EOL
Machine A	250	50	0	250	115
Machine B	150	100	200	200	150
Machine C	0	0	150	150	45
Probability	0.4	0.3	0.3	Blank	Blank

Using Software (1 of 2)

PROGRAM 3.1A QM for Windows Input for Thompson umber Example

Using Software (2 of 2)

PROGRAM 3.1B QM for Windows Output Screen for Thompson Lumber Example

Using Excel (1 of 2)

PROGRAM 3.2A Excel QM Results for Thompson Lumber Example

Using Excel (2 of 2)

PROGRAM 3.2B Key Formulas in Excel QM for Thompson Lumber Example

Decision Trees

Any problem that can be presented in a decision table can be graphically represented in a decision tree

Most beneficial when a sequence of decisions must be made

All decision trees contain decision points/nodes and state-of-nature points/nodes

At decision nodes one of several alternatives may be chosen

At state-of-nature nodes one state of nature will occur

Five Steps of Decision Tree Analysis

Define the problem

Structure or draw the decision tree

Assign probabilities to the states of nature

Estimate payoffs for each possible combination of alternatives and states of nature

Solve the problem by computing expected monetary values (EMVs) for each state of nature node

Structure of Decision Trees

Trees start from left to right

Trees represent decisions and outcomes in sequential order

Squares represent decision nodes

Circles represent states of nature nodes

Lines or branches connect the decisions nodes and the states of nature

Thompson’s Decision Tree (1 of 2)

FIGURE 3.2 Thompson’s Decision Tree

Thompson’s Decision Tree (2 of 2)

FIGURE 3.3 Completed and Solved Decision Tree for Thompson Lumber

Thompson’s Complex Decision Tree (1 of 5)

FIGURE 3.4 Larger Decision Tree with Payoffs and Probabilities for Thompson Lumber

Thompson’s Complex Decision Tree (2 of 5)

Given favorable survey results

EMV(node 2) = EMV(large plant | positive survey)

= (0.78)($190,000) + (0.22)(−$190,000) = $106,400

EMV(node 3) = EMV(small plant | positive survey)

= (0.78)($90,000) + (0.22)(−$30,000)

= $63,600

EMV for no plant = −$10,000

Thompson’s Complex Decision Tree (3 of 5)

Given negative survey results

EMV(node 4) = EMV(large plant | negative survey)

= (0.27)($190,000) + (0.73)(−$190,000)

= −$87,400

EMV(node 5) = EMV(small plant | negative survey)

= (0.27)($90,000) + (0.73)(−$30,000)

= $2,400

EMV for no plant = −$10,000

Thompson’s Complex Decision Tree (4 of 5)

Expected value of the market survey

EMV(node 1) = EMV(conduct survey)

= (0.45)($106,400) + (0.55)($2,400)

= $47,880 + $1,320 = $49,200

Expected value no market survey

EMV(node 6) = EMV(large plant)

= (0.50)($200,000) + (0.50)(−$180,000)

= $10,000

EMV(node 7) = EMV(small plant)

= (0.50)($100,000) + (0.50)(−$20,000)

= $40,000

EMV for no plant = $0

The best choice is to seek marketing information

Thompson’s Complex Decision Tree (5 of 5)

FIGURE 3.5 Thompson’s Decision Tree with EMVs Shown

Expected Value of Sample Information

Thompson wants to know the actual value of doing the survey

= (EV with SI + cost) − (EV without SI)

EVSI = ($49,200 + $10,000) − $40,000 = $19,200

Efficiency of Sample Information

Possibly many types of sample information available

Different sources can be evaluated

For Thompson

Market survey is only 32% as efficient as perfect information

Sensitivity Analysis (1 of 2)

How sensitive are the decisions to changes in the probabilities?

How sensitive is our decision to the probability of a favorable survey result?

If the probability of a favorable result (p = .45) were to change, would we make the same decision?

How much could it change before we would make a different decision?

Sensitivity Analysis (2 of 2)

p = probability of a favorable survey result

(1−p) = probability of a negative survey result

EMV(node 1) = ($106,400)p +($2,400)(1−p)

= $104,000p + $2,400

We are indifferent when the EMV of node 1 is the same as the EMV of not conducting the survey

$104,000p + $2,400 = $40,000

$104,000p = $37,600

p = $37,600÷$104,000 = 0.36

If p < 0.36, do not conduct the survey

If p > 0.36, conduct the survey

End of Lesson 4

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Efficiency of sample information = 100%

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19,200

Efficiency of sample information = 100%

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60,000